How to Serialize Binary Tree

Question

I went to an interview today where I was asked to serialize a binary tree. I implemented an array-based approach where the children of node i (numbering in level-order trave

Answer 1

All those articles talk mostly about the serialization part. The deserialization part is slightly tricky to do in one pass.

I have implemented an efficient solution for deserialization too.

Problem: Serialize and Deserialize a binary tree containing positive numbers.

Serialization part:

1. Use 0 to represent null.
2. Serialize to list of integers using preorder traversal.

Deserialization part:

1. Takes in list of integers and uses recursive helper method for deserialization.
2. Recursive deserializer returns a pair (BTNode node, int nextIndexToRead) where node is tree node constructed so far, and nextIndexToRead is position of next number to be read in the serialized list of numbers.

Below is the code in Java:

public final class BinaryTreeSerializer
{
    public static List Serialize(BTNode root)
    {
        List serializedNums = new ArrayList();

        SerializeRecursively(root, serializedNums);

        return serializedNums;
    }

    private static void SerializeRecursively(BTNode node, List nums)
    {
        if (node == null)
        {
            nums.add(0);
            return;
        }

        nums.add(node.data);
        SerializeRecursively(node.left, nums);
        SerializeRecursively(node.right, nums);
    }

    public static BTNode Deserialize(List serializedNums)
    {
        Pair pair = DeserializeRecursively(serializedNums, 0);

        return pair.node;
    }

    private static Pair DeserializeRecursively(List serializedNums, int start)
    {        
        int num = serializedNums.get(start);

        if (num == 0)
        {
            return new Pair(null, start + 1);
        }

        BTNode node = new BTNode(num);

        Pair p1 = DeserializeRecursively(serializedNums, start + 1);
        node.left = p1.node;

        Pair p2 = DeserializeRecursively(serializedNums, p1.startIndex);
        node.right = p2.node;

        return new Pair(node, p2.startIndex);
    }

    private static final class Pair
    {
        BTNode node;
        int startIndex;

        private Pair(BTNode node, int index)
        {
            this.node = node;
            this.startIndex = index;
        }
    }
}

public class BTNode 
{
    public int data;
    public BTNode left;
    public BTNode right;

    public BTNode(int data)
    {
        this.data = data;
    }
}